(a) In the circuit shown, calculate

(i) the `p.d.` between `B` and `D`
(ii) the `p.d.` across the terminals of each of the cells `G` and `H`.
(b) Find the potentials of `A, B, C` and `D` and current through `1 Omega` and `2 Omega` resistance.
.

Explanation

(a)
(i) Cells in series, equivalent emf `= 2 - 1 = 1 V`
(ii) Cells in series, equivalent emf `= 3 - 1 = 2 V`
Net resistance `=1 + 3 = 4 Omega`

`E = ((2)/(4) -(1)/(3))/((1)/(4) +(1)/(3)) = (2)/(7) V`
`r =(3 xx 4)/(3 + 4) = (12)/(7) Omega`
`i = (2//7)/((12)/(7) +2) = (1)/(13) A`
(i) `V_B - V_D = i xx 2= (2)/(13) V`
(ii) Cell of emf `2 V`
`V_B - V_D = 2 0 i_0 xx 4`
`(2)/(13) = 2 - 4 i_0 rArr i_0 = (6)/(13) A`
Cell `H` : (taking current)
`V_E - V_B = 1 + i_0 xx 1 = 1 + (6)/(13) xx 1 = (19)/(13) V`
Alternatively by Kirchhoff's rules

`Lo op ①, D to D, path DEBD`
`V_D - 3 i_1 + 3 - 1 - i_1 + 2 i_2 = V_D`
`4 i_1 - 2 i_2 = 2`
`2 i_1 - i_2 = 1`...(i)
`Lo op ②, D to D, path DBFD`
`V_D - 2 i_2 - 2(i_1 + i_2) + 2 -1-(i_1 + i_2) = V_D`
`-3 i_1 - 5 i_2 + 1 = 0`
`3 i_1 + 5 i_2 = 1`...(ii)
Solving (i) and (ii), `i_1 = (6)/(13) A, i_2 = -(1)/(13) A`
Direction of `i_2` is opposite
Cell `G` : (supplying current)
`V_E - V_d = 3 - 3 i_1 = 3 - 3 xx (6)/(13) = (21)/(13) V`
Cell `H` : (taking current)
`V_E - V_B = 1 + i_1 = 1 + (6)/(13) = (19)/(13) V`
(b)
The point `A` is attached to ground i.e., `V_A = 0`
by branch `CA, V_C - V_A = 5`
`V_C - 0 = 5`
`V_C = 5 V`
By branch `ADC, V_A + 2 i_1 - 10 = V_C`
`2 i_1 - 10 = 5`
`i_1 = 7.5 A`
By branch `ABC, V_C - i_2 xx 1 - 2 = V_A`
`5 = i_2 + 2`
`i_2 = 3 A`
`V_B - V_A = 2`
`V_B - 0 = 2 rArr V_B = 2 V`
`V_D - V_A = 2 i_1 = 15`
`V_D - 0 = 15 rArr V_D = 15 V`
`V_A = 0, V_B = 2 V, V_c = 5 V, V_D = 15 V`
Current in `1 Omega : 3 A`
Current in `2 Omega : 7.5 A`.